# Anti-unitary operator and hamiltonian

For a symmetry represented by a unitary operator $$U$$ to be a dynamical symmetry, we require the condition that $$Ue^{(-iHt/\hbar)}=e^{(-iHt/\hbar)}U$$ which implies $$UHU^*=H$$.

If instead $$U$$ is an anti-unitary opertor, show that the above equation would imply that $$UHU^*=-H$$.

I'm not too sure how to do this question. I don't really understand how the first implication is derived from the condition, and secondly I don't see how this changes for an anti-unitary operator. $$H$$ is the Hamiltonian, and the definitions of unitary operator and anti-unitary operators are as follows:

A unitary operator $$U$$ on a Hilbert space is a linear map $$U :\mathcal{H} \rightarrow \mathcal{H}$$ that obeys $$UU^*=U^*U=1_{\mathcal{H}}$$ ($$U^*$$ being the adjoint).

An anti-unitary operator on a hilbert space is a surjective linear map $$A :\mathcal{H} \rightarrow \mathcal{H}$$ obeying $$\langle A\phi |A\psi \rangle = \overline {\langle \phi | \psi \rangle} = \langle \psi | \phi \rangle$$

• You missed the definition of adjoint operator for antilinear operators. Usually, it is a source of disasters. A much better pair of statements (equivalent to your pair) would be $UHU^{-1} = H$ is $U$ is unitary and $UHU^{-1} = -H$ if $U$ is antiunitary. Feb 9 at 17:36
• @ValterMoretti I am unsure what you mean sorry Feb 9 at 17:39

A unitary operator is a linear surjective operator $$U : {\cal H} \to {\cal H}$$ that preserves the norm. It is equivalent to $$U^*=U^{-1}$$, namely $$UU^*=U^*U=I$$, where $$U^*$$ henceforth denotes the adjoint of $$U$$.

An antiunitary operator is an antilinear surjective operator $$U : H \to H$$ that preserves the norm. It is equivalent to $$U$$ bijective such that $$\langle U\psi|\phi\rangle = \overline{\langle \psi| U\phi\rangle}\:,\quad \forall \psi, \phi \in {\cal H}\:.$$ Now suppose that, in either cases, for all $$t\in \mathbb{R}$$ $$U e^{-itH} = e^{-itH}U\:.$$ By applying $$U^{-1}$$ on the right, we get the equivalent condition $$Ue^{-itH} U^{-1}= e^{-itH}\:.\tag{1}$$ From spectral calculus or other more elementary procedures, e.g., expanding the exponential as a series if $$H$$ is bounded and paying attention to $$U i H = -iUH$$ in view if antilinearity of $$U$$ if it is the case, (1) entails $$e^{\mp itUHU^{-1}} = e^{-itH}\:.$$ Computing the derivative at $$t=0$$ (Stone's theorem) of both sides (on the relevant dense domain of $$H$$ which turns out to be invariant under $$U^{-1}$$, directly form the uniqueness part of Stone's theorem): $$\pm UHU^{-1} = H\:,$$ that is $$UHU^{-1} = \pm H\:,\tag{2}$$ where the sign $$-$$ is reserved to the antiunitary case. In case of a unitary oparator, we have also found that $$UHU^{*} = H$$ because $$U^*=U^{-1}$$. In case of an antiunitary $$U$$, with a suitable definition ($$\dagger$$) of adjoint operator for antilinear operators, we can equivalently rewrite (2) as $$UHU^{*} = -H\:.$$

However the definition of adjoint of an antiunitary operator is usually delicate and, to my personal experience, it is a source of mistakes. Dealing with symmetries it is much better to use $$U^{-1}$$ in both cases in place of $$U^*$$.

$$(\dagger)$$ $$\langle \psi|A \phi\rangle = \overline{\langle A^*\psi| \phi\rangle}$$ for all $$\psi,\phi\in {\cal H}$$ assuming $$A$$ everywhere defined and antilinear.

• Thank you for the detailed answer! Feb 10 at 12:05

There are a couple of confusing (or even wrong?) points in the post. First, I assume $$U^*$$ means $$U^\dagger$$, the adjoint of $$U$$. A unitary symmetry means $$UHU^\dagger=H$$.

An anti-unitary operator is first of all an anti-linear operator instead of a linear one. If $$U$$ is anti-unitary symmetry, then one still has $$UHU^\dagger=H$$, there should not be an extra minus sign. However, the definition of adjoint for anti-linear operator is different from that of a linear operator.

Edit: the other answer is correct. Usually for a time-reversal symmetry (which is the most common way one gets anti-unitary symmetry) we also take $$t$$ to $$-t$$ so $$UHU^\dagger=H$$. But if $$U$$ is just anti-unitary without $$t$$ going to $$-t$$, then because $$Ui=-iU$$ we have the extra minus sign.

• Can whoever voting this down comment on the reason? Feb 9 at 19:00